Permutations of squares

Question

Consider the following problem: Imagine you choose 3 squares out of a 3x3 field. So you can have a line upwords or downwords, two squares downwords and one to the right etc. So let's number every square (see MWE). Now it is just a mathematicaly problem: $binom(9,3)$.

But since the squares all look the same, there is no difference between (1,2,3) and (3,2,1) and so on. So when we take a look here it is a pretty good solution, but we got the problem with the duplicates.

I thought of writing a python script which exoprts the correct permuations into a .csv file and then TikZ or pgfplots(maybe it can be done there, too) can read the file and solve the "square problem".

So my question is: How can I "cross out" the duplicates in order to get sth. like in the MWE?

\documentclass[border=5pt,tikz]{standalone}
\newcommand{\setcircle}[3]{
    \pgfmathsetmacro\testnum{int(mod(#1,3))}
    \ifnum\testnum=0
        \pgfmathsetmacro\oxpos{3}
        \pgfmathsetmacro\oypos{floor(#1/3)-1}
    \else
        \pgfmathsetmacro\oxpos{mod(#1,3)}
        \pgfmathsetmacro\oypos{floor(#1/3)}
    \fi

    \pgfmathsetmacro\testnum{int(mod(#2,3))}
    \ifnum\testnum=0
        \pgfmathsetmacro\txpos{3}
        \pgfmathsetmacro\typos{floor(#2/3)-1}
    \else
        \pgfmathsetmacro\txpos{mod(#2,3)}
        \pgfmathsetmacro\typos{floor(#2/3)}
    \fi

    \pgfmathsetmacro\testnum{int(mod(#3,3))}
    \ifnum\testnum=0
        \pgfmathsetmacro\thxpos{3}
        \pgfmathsetmacro\thypos{floor(#3/3)-1}
    \else
        \pgfmathsetmacro\thxpos{mod(#3,3)}
        \pgfmathsetmacro\thypos{floor(#3/3)}
    \fi

    \draw (\oxpos,-\oypos) circle(.5) node {#1};
    \draw (\txpos,-\typos) circle(.5) node {#2};
    \draw (\thxpos,-\thypos) circle(.5) node {#3};
}
\begin{document}
    \begin{tikzpicture}
        \foreach \x in {1,...,9}
        {
            \pgfmathsetmacro\testnum{int(mod(\x,3))}
            \ifnum\testnum=0
                \pgfmathsetmacro\xpos{3}
                \pgfmathsetmacro\ypos{floor(\x/3)-1}
            \else
                \pgfmathsetmacro\xpos{mod(\x,3)}
                \pgfmathsetmacro\ypos{floor(\x/3)}
            \fi

            \draw (\xpos,-\ypos) circle(.5) node {\x};
        }

        \begin{scope}[xshift=-3cm,yshift=-4cm]
            \setcircle{1}{2}{5}
        \end{scope}

        \begin{scope}[yshift=-4cm]
            \setcircle{1}{2}{3}
        \end{scope}

        \begin{scope}[xshift=3cm,yshift=-4cm]
            \setcircle{3}{6}{8}
        \end{scope}
    \end{tikzpicture}
\end{document}

Output:

Answer 1

This draws all the inequivalent combinations.

\documentclass[tikz,border=3.14mm]{standalone}
\newcounter{mystep}
\begin{document}
\begin{tikzpicture}[insert circle/.style={insert path={%
({mod(#1-1,3)*0.75},{int((#1-1)/3)*0.75}) node[circle,draw]{#1}}}]
\foreach \X [evaluate=\X as \Ymin using {int(\X+1)}] in {1,...,9}
{\foreach \Y [evaluate=\Y as \Zmin using {int(\X+1)}]in {\Ymin,...,9}
{\foreach \Z in {\Zmin,...,9}
{\ifnum\X<\Y
   \ifnum\Y<\Z
     \stepcounter{mystep}
     \begin{scope}[xshift={mod(\number\value{mystep}-1,7)*3cm},
     yshift={-int((\number\value{mystep}-1)/7)*3cm}]
      \path[insert circle/.list={\X,\Y,\Z}];
     \end{scope}
   \fi
 \fi  
}}}
\typeout{\number\value{mystep}\space combinations}
\end{tikzpicture}
\end{document}